Question

How many different committees can be formed from 12 teachers and 43 students if the committee consists of 3 teachers and 4 ​students?
The committee of 7 members can be selected in BLANK
different ways.​

Answer 1

Answer:

$\displaystyle 27150200$

Step-by-step explanation:

we are two conditions

• committees can be formed from 12 teachers and 43 students
• the committee consists of 3 teachers and 4 students

In choosing a committee, order doesn't matter; in case of teachers we need the number of combinations of 3 people chosen from 12

remember that,

$\displaystyle\binom{n}{r} = \frac{n!}{r!(n - r)!}$

with the condition we obtain that,

• $n = 12$
• $r = 3$

therefore substitute:

$\displaystyle\binom{12}{3} = \frac{12!}{3!(12 - 3)!}$

simplify Parentheses:

$\displaystyle\binom{12}{3} = \frac{12!}{3! \cdot9!}$

rewrite:

$\rm \displaystyle\binom{12}{3} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(1 \times 2 \times 3 )\cdot1 \times 2 \times 3 \times 4 \times 5 \times 6 \times 7 \times 8 \times 9}$

reduce fraction:

$\rm \displaystyle\binom{12}{3} = \frac{12 \times 11 \times 10}{1 \times 2 \times 3 }$

rewrite 12 and 10:

$\rm \displaystyle\binom{12}{3} = \frac{3 \times 2 \times 2 \times 11 \times 10}{1 \times 2 \times 3 }$

reduce fraction:

$\rm \displaystyle\binom{12}{3} = 2 \times 11 \times 10$

simplify multiplication:

$\rm \displaystyle\binom{12}{3} = 220$

In case of students we need the number of combinations of 4 students choosen from 43 therefore,

$\displaystyle\binom{43}{4} = \frac{43!}{4!(43 - 4)!}$

simplify which yields:

$\displaystyle\binom{43}{4} = 123410$

hence,

The committee of 7 members can be selected in BLANK different ways is

$\displaystyle 123410 \times 220$

$\displaystyle \boxed{27150200}$

and we're done!